Baire Spaces

Updated 31 Aug 2026Chapter (PDF)

We only cover Baire spaces in this section, since the other two sections are unimportant to us currently.

Let's first consider what properties a set with empty interior has (that is, a set whose interior is the empty set). Many consequences come fairly quickly from this. For example, since the interior of a set $A$ is the union of all open subsets of $A$ , we conclude that a set with empty interior contains no open set besides the empty set. As another consequence, every open neighborhood about $x \in A$ must intersect $X \setminus A$ , and so $X \setminus A$ is dense in $X$ . There are two perspectives to be taken when considering a set with empty interior: the set itself containing no open set, and the rest of the space being dense.

[8.0.1]Definition(Empty Interior)#

A set $A$ is said to have empty interior based on the following three equivalent characterizations.

$\Int(A) = \emptyset$ .
The only open set $A$ contains is the empty set.
$X \setminus A$ is dense.

[8.0.2]Definition(Baire Space)#

A space $X$ is said to be a Baire space if, for any countable collection $\{A_n\}$ of closed sets with empty interior, we have that $\bigcup A_n$ also has empty interior.

[8.0.3]Theorem(Characterization of Baire Space)#

A space $X$ is a Baire space if and only if, for any countable collection $\{U_n\}$ of open dense sets, we have that $\bigcap A_n$ is also dense.

[8.0.4]Theorem(Baire Category Theorem)#

If $X$ is a complete metric or compact Hausdorff space, then it is a Baire space.

Proof.

Let $\{A_n\}$ be a countable collection of closed sets with empty interior. We show $\bigcup A_n$ also has closed interior. That is, for any nonempty set $U_0$ in $X$ , we show there is some $x \in U_0$ such that $x \notin \bigcup A_n$ .

Since $A_0$ has empty interior, note that $A_0$ cannot contain $U_0$ , and so there is some $x \in U_0$ such that $x \notin A_0$ . Note $X$ is regular and $A_0$ is closed, so there exists an open neighborhood $U_1$ of $x$ such that

\bar{U_1} \cap A_0 = \emptyset,

\bar{U_1} \subseteq U_0,

\diam(U_1) < 1 \quad \text{(if $X$ is metric)}.

We can continue this process by induction: any $A_k$ cannot contain $U_k$ , and so there is some $x \in U_k$ such that $x \notin A_k$ . Then we can choose an open neighborhood $U_{k + 1}$ such that

\bar{U_{k + 1}} \cap A_k = \emptyset,

\bar{U_{k + 1}} \subseteq U_k,

\diam(U_{k + 1}) < 1/(k + 1) \quad \text{(if $X$ is metric)}.

We show $\bigcap \bar{U_n}$ is nonempty. If there is some $x \in \bigcap \bar{U_n}$ , then we have that $x \in \bar{U_1} \subseteq U_0$ and $x \notin A_n$ for every $n$ , completing the proof. In the case of $X$ being compact Hausdorff, we note that the decreasing (ordered by inclusion) sequence of closed sets $\{\bar{U_n}\}$ has the finite intersection property, and so by compactness, $\bigcap \bar{U_n}$ is nonempty.

This result holds for complete metric spaces too. Indeed, since $\diam(U_n) \to 0$ , we may form a sequence $(x_n)$ such that $x_k \in C_k$ and note that it is Cauchy. By completeness, it converges to some $x$ such that $x \in \bar{C_k}$ for every $k$ , and so $\bigcap \bar{C_k}$ is nonempty.

[8.0.5]Theorem#

Any open subset of a Baire space is a Baire space.

[8.0.6]Theorem#

Let $X$ be a space and $(Y, d)$ a metric space. Suppose $f_n: X \to Y$ is a sequence of functions such that $f_n \to f$ pointwise. If $X$ is Baire, then the set of points that $f$ is continuous on is a dense set in $X$ .